GROUP THEORY – 3 (CYCLIC GROUPS – LAGRANGE’S THEOREM) SOME DEFINITIONS, THEOREMS AND SOME PROPERTIES: Definition: A group G is said to be a cyclic group if G can be written as G = {an: n }. In additive notation a cyclic group G is written as {na: n }. This a is called a generator of the group. Definition: Let G be a group and H be a subgroup of G. For any element aG we define a left coset aH (right coset Ha) of H in G by aH = {ah: h H}, (Ha = {ha : h H}). Theorem 1: Every subgroup of a cyclic group is cyclic. Theorem 2: A group of prime order is cyclic. Theorem 3: If a is a generator of a cyclic group G, then a-1 is also a generator of G. Theorem 4: Every cyclic group is abelian. Theorem 5: Let G be a cyclic group generated by a. Then o(G) = n if and only if o(a) = n. Theorem 6: Let G be a cyclic group generated by a. Then g is infinite if and only if o(a) is infinite. Theorem 7: A finite group G of order n is cyclic if and only if there is some element b in G of order n such that o(b) = n. Theorem 8: Let G be a group of order n generated by a, then ar is also a generator of G if and only if r is less than n and prime to n. Corollary: A finite cyclic group G of order n has φ(n) generators. Theorem 9: Every non-trivial subgroup of an infinite cyclic group is infinite. Theorem 10: A finite cyclic group G of order n has a subgroup of order d for each divisor d of n. Theorem 11: Let H be a subgroup of a group G, then hH = H if and only if h H. Theorem 12: Let H be a subgroup of a group G, then any two left cosets of H in G are either disjoint or identical. Theorem 13: Let H be a subgroup of a group G and a,b G, then the left cosets aH and bH are identical if and only if a-1b H. Theorem 14: Let G be a group and H be a subgroup of G, then any two left cosets of H in G have the same cardinality. Theorem (Lagrange)15: The order of any subgroup of a finite group G divides the order of the group G. 1. G is a nontrivial group with no nontrivial proper subgroup. Show that G is a group of prime order . SOLUTION: Let G be a non-trivial group with identity element e. Let a be a non-identity element of G. Let us consider the cyclic subgroup H generated by a, i.e H = <a>. Since the group G has no proper subgroup so G = H. If G is an infinite cyclic group generated by a then <a2> is a cyclic subgroup generated by a2 and it is proper. So G can not be infinite. Therefore G is a finite group. Let G be a group of order n . If n is composite then n = pq for some integers p and q. G = <a> and o(G) = pq so o(a) = pq and hence o(ap) =q, i.e order of the cyclic group <ap> is q and it is also a proper subgroup of G. Hence G can not be a group of composite order. Therefore G is a group of prime order. 2. If an abelian group G of order 10 contains a subgroup of order 5, then show that G must be cyclic. SOLUTION: Let G be a group of order 10 containing a subgroup H of order 5. Since H is a group of prime order so H is cyclic. H = <a> . So o(a) = 5. If all elements are of order 5, then the group K = <b> where b G-H is also cyclic and is of order 5. Then HKis a subgroup of H as well as of K. So by Lagrange’s Theorem o( ) is a divisor of o(H) as well as of o(K) . If o( ) = 5, then =H =K. So o( ) = 1. Hence o(HK) = o(H)o(K)/o( ) = 5.5/1 = 25 but HK is a subgroup of G and o(G) = 10. Therefore each element of G- H is of order 2. Let b G-H. So o(b) = 2. Since G is commutative and gcd(2,5) = 1 so o(ab) = o(a)o(b) = 2.5 =10. Hence G is cyclic. 3. If G be a group of order pq where p,q are distinct primes, show that every proper subgroup of G is cyclic. SOLUTION: Same as problem no. 2. 4. Prove that all the proper subgroups of a group G of order 8 are commutative. SOLUTION: The orders of subgroups of a group G of order 8 are divisors of 8, i.e 1,2,4,8. So the orders of proper subgroups of G are 1,2,4. If H be a subgroup of order 1, then the group is obviously commutative. If H is of order 2 then H is a group of prime order hence H is cyclic, so H is commutative. If H is of order 4, then each non- identity element of H is of order 2 or 4, (divisors of 4). If there exists an element a of order 4 then H is cyclic and hence H is commutative. If each non-identity element is of order 2 then for any two elements a and b, o(a) = o(b) = o(ab) = 2, i.e a2 = b2 =(ab)2= e, where e is the identity element of G. So a = a-1, b = b-1 , (ab) = (ab)-1 = b-1a-1 = ba, i.e ab = ba. So H is commutative. Hence each proper subgroup of G is commutative. 5. G is a non-commutative group of order 10. Show that G must have a subgroup of order 5. SOLUTION: Let G be a non-commutative group of order 10. The order of each element of G is a divisor of 10 i.e 1,2,5 or 10. If there is some element of order 10 , then the group is cyclic and hence commutative but the group is non-commutative, a contradiction arises. So there is no element of order 10. If each non-identity element is of order 2, then a2 = e for all aG where e is the identity element of G. a2 = eaa1 for all aG . Then ab = (ab)-1 = b-1a-1 = ba, i.e G is commutative. Again we have arrived at a contradiction. So each non-identity element of G is not of order 2. Hence there is some element of order 5 and a subgroup of order 5. 6. G is a group of order 27. Prove that G must have a subgroup of order 3. SOLUTION: Let G be a group of order 27. The orders of elements of G are the divisors of order of G, i.e 1,3,9,27. If G contains an element of order 27, then the group is cyclic. Let G be generated by a and o(a) = 27, i.e o(a9) = 3. So the group <a9> is a subgroup of order 3. If there is some element b of order 9, then o(b3) = 3 and the subgroup <b3> is a subgroup of order 3. If there is some element c of order 3 then the group <c> is a subgroup of order 3. So in any case G contains a subgroup of order 3. 7. G is a non-abelian group of order 8. Prove that G must have an element of order 4. SOLUTION: Left as an exercise. 8. Let G be a nonabelian group of order 2p , p odd prime. Find the number of elements of order p. SOLUTION: Left as an exercise. .